many species of cuckoos are brood parasites. The females lay their eggs in the nests if similar bird species that then raise the young cuckoos at the expense of their own young. The question might be asked, "Do the cuckoos lay the same size eggs regardless of the size of the bird whose nest they use"? Data on the lengths, in millimeters of cuckoo eggs found in the nests of six bird species- Meadow Pipit, Tree pipit, Hedge Sparrow, Robin, Pied Wagtail and wren are provided on the WeissStats site. These data are collected by the late O. Latter in \(1902\) and used by L. Tippett in his text The Methods of Statistics.

a. Obtain individual normal probability plots and the standard deviations of the sample.

b. Perform a residual analysis

c. use your results from part (a) and (b) to decide whether conducting a one-way ANOVA test on the data is reasonable. If so, also do parts (d) and (e).

d. use a one-way ANOVA test to decide, at the\(5%\) significance level whether, the data provide sufficient evidence to conclude that a difference exists among the means of the populations from which the samples were taken.

e. Interpret your results from part (d).

The sample data is given into the paper

Let’s take the random sample of length \(120\0.

Draw a normal probability plot using function “normplot” in MATLAB.

Program:

Query:

• Then generate the normal probability plot.

Let’s take the random sample of length \(120\).

Then calculate the residual using relation

\(residual = data -fit\)

Then, draw a normal probability plot using function “normplot” in MATLAB.

Program:

Query:

• Then generate the normal probability plot of the residuals.

Calculate the SST, SSTR and SSE using given relation

\(SST=\sum x^{2}-\frac{(\sum x)^{2}}{n}\)

\(SST=474-\frac{(80)^{2}}{16}=7.4\)

\(SSTR=\frac{\sum (x_{i})^{2}}{n_{i}}-\frac{\sum (x)^{2}}{n}\)

\(SSTR=\frac{12^{2}}{3}+\frac{25^{2}}{5}+\frac{15^{2}}{5}+\frac{18^{2}}{3} -\frac{(80)^{2}}{16}=46\)

\(SSE=SST-SSTR=28\)

Then,

\(df_{T}=k-1=3-1=2\)

\(df_{E}=n-k=10-3=7\)

\(MSTR=\frac{SSTR}{df_{T}}=\frac{24}{2}=12\)

\(MSE=\frac{SSE}{df_{E}}=\frac{16}{7}=2.2857\)

\(F=\frac{MSTR}{MSE}=\frac{12}{2.2857}\approx 5.25\)

After calculating these values put all into the table and get ANOVA table

The \(p-\)value is the probability value which obtaining by the test statistics, or a value more extreme. The \(P-\)value is the number in the row title of the \(F-\)distribution table which containing \(F-\)value in the row \(dfd=df_{E}=7\) and in the column \(dfn=df_{T}=2\)

So, the \(p-\)value lie between

\(0.025<P<0.050\)

And if the \(p-\)value is less than significance level then it will reject the null hypothesis.

\(P<0.05\Rightarrow\) Reject \(H_{0}\)